Problem 3
Question
In Problems, graph the curve traced by the given vector function. \(\mathbf{r}(t)=t \mathbf{i}+2 t \mathbf{j}+\cos t \mathbf{k} ; t \geq 0\)
Step-by-Step Solution
Verified Answer
The curve is a helix spiralling upwards along the xz-plane.
1Step 1: Identify the Components
The given vector function is \( \mathbf{r}(t) = t \mathbf{i} + 2t \mathbf{j} + \cos t \mathbf{k} \). This means the \(x\)-component is \(t\), the \(y\)-component is \(2t\), and the \(z\)-component is \(\cos t\).
2Step 2: Analyze Each Component
As \(t\) increases, the \(x\)-component \(x = t\) increases linearly. Similarly, the \(y\)-component \(y = 2t\) also increases linearly, but twice as fast as the \(x\)-component. The \(z\)-component \(z = \cos t\) oscillates between -1 and 1.
3Step 3: Determine the Path of Movement
The \(x\) and \(y\) positions increase with \(t\), meaning the graph spirals upwards along the xy-plane. Since \(z = \cos t\) oscillates, it causes the curve to move up and down as it spirals.
4Step 4: Sketch the Curve
Begin by plotting points with increasing \(t\) on the xyz-plane. Note that as \(t\) increases, the path spirals upwards along the x-axis while oscillating in the z-direction due to the cosine term. The overall shape is a helical curve winding upwards.
Key Concepts
Parametric EquationsHelical CurvesVector Components
Parametric Equations
A parametric equation represents a method for defining a curve or a path in space using one or more parameters, usually denoted as \(t\). Instead of providing a direct relation between \(x\), \(y\), and \(z\), this approach uses separate equations to express each coordinate. For instance, in the given vector function \( \mathbf{r}(t) = t \mathbf{i} + 2t \mathbf{j} + \cos t \mathbf{k} \), the parameter \(t\) determines the values of \(x\), \(y\), and \(z\) independently.
Parametric equations are powerful because they allow us to describe complex curves easily. The main advantage is the flexibility they offer, enabling us to model paths that would be difficult to represent with traditional equations. In the vector function given, each component depends directly on \(t\), offering a clear and structured way to understand how a point moves through space.
To visualize these equations, we often set a range for \(t\) and compute each component, mapping out the curve on a 3D graph.
Parametric equations are powerful because they allow us to describe complex curves easily. The main advantage is the flexibility they offer, enabling us to model paths that would be difficult to represent with traditional equations. In the vector function given, each component depends directly on \(t\), offering a clear and structured way to understand how a point moves through space.
To visualize these equations, we often set a range for \(t\) and compute each component, mapping out the curve on a 3D graph.
Helical Curves
Helical curves, commonly referred to as helices, are fascinating structures similar to the shape of a spring or spiral staircase. These curves have a special property: as one moves along the helix, the path typically rotates while also advancing in a direction, creating a three-dimensional spiraling movement.
In the example from the exercise, the vector function \( \mathbf{r}(t) = t \mathbf{i} + 2t \mathbf{j} + \cos t \mathbf{k} \) describes a helical path. The linear functions \(x = t\) and \(y = 2t\) indicate a steady upward spiral in the \(xy\)-plane, while the oscillating function \(z = \cos t\) adds an oscillatory motion in the \(z\)-direction.
To understand this better, envision a point moving along the \(xy\)-plane, consistently shifting upwards because both \(x\) and \(y\) increase linearly. Simultaneously, thanks to \(\cos t\), the point moves up and down in the \(z\)-axis as it spirals. This dynamic interplay of linear and oscillatory motion results in the helix's unique 3D form.
In the example from the exercise, the vector function \( \mathbf{r}(t) = t \mathbf{i} + 2t \mathbf{j} + \cos t \mathbf{k} \) describes a helical path. The linear functions \(x = t\) and \(y = 2t\) indicate a steady upward spiral in the \(xy\)-plane, while the oscillating function \(z = \cos t\) adds an oscillatory motion in the \(z\)-direction.
To understand this better, envision a point moving along the \(xy\)-plane, consistently shifting upwards because both \(x\) and \(y\) increase linearly. Simultaneously, thanks to \(\cos t\), the point moves up and down in the \(z\)-axis as it spirals. This dynamic interplay of linear and oscillatory motion results in the helix's unique 3D form.
Vector Components
Vector components play a foundational role in vector calculus, forming the building blocks for determining how forces or paths behave across space. Each component corresponds to one of the fundamental axes - typically \(x\), \(y\), and \(z\) - and assigns a specific function to describe movement or position changes along that axis.
In the vector function \( \mathbf{r}(t) = t \mathbf{i} + 2t \mathbf{j} + \cos t \mathbf{k} \), the components are \(x = t\), \(y = 2t\), and \(z = \cos t\). Each is responsible for defining part of the overall motion.
- The \(x\)-component \(t\) implies a linear increase along the \(x\)-axis, showing constant progression.- The \(y\)-component \(2t\) also signifies linear movement, but with a steeper gradient, doubling the speed compared to \(x\).- The \(z\)-component \(\cos t\) is different as it provides an oscillating movement, confining the path upward and downward within the range of -1 to 1 periodically.
With these vector components combined, they create a comprehensive picture of a particle's trajectory, showing how it changes position over time in a complex yet predictable manner.
In the vector function \( \mathbf{r}(t) = t \mathbf{i} + 2t \mathbf{j} + \cos t \mathbf{k} \), the components are \(x = t\), \(y = 2t\), and \(z = \cos t\). Each is responsible for defining part of the overall motion.
- The \(x\)-component \(t\) implies a linear increase along the \(x\)-axis, showing constant progression.- The \(y\)-component \(2t\) also signifies linear movement, but with a steeper gradient, doubling the speed compared to \(x\).- The \(z\)-component \(\cos t\) is different as it provides an oscillating movement, confining the path upward and downward within the range of -1 to 1 periodically.
With these vector components combined, they create a comprehensive picture of a particle's trajectory, showing how it changes position over time in a complex yet predictable manner.
Other exercises in this chapter
Problem 3
Sketch the level curve or surface passing through the indicated point. Sketch the gradient at the point. $$ f(x, y)=y-x^{2} ;(2,5) $$
View solution Problem 3
In Problems, compute the gradient for the given function. $$ F(x, y, z)=\frac{x y^{2}}{z^{3}} $$
View solution Problem 3
Fill in the blank or answer true/false. Where appropriate, assume continuity of \(P, O\), and their first partial derivatives. The binormal vector is perpendicu
View solution Problem 3
Use the divergence theorem to find the outward flux \(\iint_{S}(\mathbf{F} \cdot \mathbf{n}) d S\) of the given vector field \(\mathbf{F}\). \(\mathbf{F}=x^{3}
View solution